Q.Find the equations of the tangents to the hyperbola $x^2 - 4y^2 = 4$ which are

Use $c^2=a^2m^2-b^2$ for a line $y=mx+c$ tangent to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$; check part (i) carefully since its slope coincides with the asymptote slope.

Write the hyperbola in standard form: $x^2-4y^2=4 \implies \dfrac{x^2}{4}-\dfrac{y^2}{1}=1$, so $a^2=4,\,b^2=1$. Its asymptotes are $y=\pm\dfrac{b}{a}x=\pm\dfrac12x$.

The line $x+2y=0$, i.e. $y=-\dfrac12x$, has slope $-\dfrac12$.

(i) Parallel to $x+2y=0$ (slope $m=-\tfrac12$): Note this slope is exactly the hyperbola's asymptote slope. A tangent to the hyperbola at a point $(x_1,y_1)$, namely $\dfrac{xx_1}{4}-yy_1=1$, has slope $\dfrac{x_1}{4y_1}$. Setting this equal to $-\dfrac12$ gives $x_1=-2y_1$. But $(x_1,y_1)$ must also lie on the hyperbola: $x_1^2-4y_1^2=4 \implies (-2y_1)^2-4y_1^2=4 \implies 4y_1^2-4y_1^2=4 \implies 0=4$, which is impossible. So no point on the hyperbola has a tangent of slope $-\tfrac12$ — a line parallel to an asymptote can never be tangent to a hyperbola. Applying the formula $c^2=a^2m^2-b^2$ mechanically gives $c^2=4\left(\tfrac14\right)-1=0$, i.e. $c=0$, but the line $y=-\tfrac12x$ (i.e. $x+2y=0$) is exactly the asymptote itself and does not touch the hyperbola at any finite point — substituting $x=-2y$ into $x^2-4y^2=4$ gives $4y^2-4y^2=0\ne4$, confirming it never meets the curve. So …